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3 years ago

In this version of golf the winner is the golfer who wins the greater number of holes. 1.miniature golf

Physics

2 answers:

Veseljchak [2.6K]3 years ago

8 0

It’s gonna be tournament play

professor190 [17]3 years ago

8 0

Answer:

Tournament play

Explanation:

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The wheels of a car have radius 12 in. and are rotating at 600 rpm. Find the speed of the car in mi/h.

guajiro [1.7K]

Answer:

21.4 mph

Explanation:

Circumference of tire in FEET = pi * d = pi * 1 ft = pi feet

pi feet x 600 rot/min * 60 min /hr * 1 mile / 5280 feet = 21.4 mph

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1 year ago

Why is it advisable to wear long sleeves when a student works in a chemistry lab

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So then if they do spill the chemical then it gets on their cloths and not on then it dosen"t harm them instead it ruins their shirt.

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3 years ago

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A manganese atom is pictured below.

Rufina [12.5K]

Manganese has 2 (two) electron that would free floating and able to form a metallic bond.
The electronic configuration of manganese is (Ar) 3d5 4s2. The two electron in 4s orbital are the valence electron which can freely move from one place to another.

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3 years ago

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$